Probability of finding a particle in the right half of a rectangular potential well

[keyzan]

Homework Statement

1.Determine the probability of finding the particle in the right half of the hole at $t=0$.

Relevant Equations

$|\psi \rangle = \frac{1}{\sqrt{2}}[|1\rangle + |2\rangle]$

Hi guys it's me again. I need help with this exercise which reads: a particle of mass m, placed in an infinite rectangular one-dimensional potential well that confines it in the segment between $x = -\frac{a}{2} and x=\frac{a}{2}$, is at instant $t=0$ in the state:

$|\psi \rangle = \frac{1}{\sqrt{2}}[|1\rangle + |2\rangle]$

Where kets 1 and 2 are normalized and represent the ground and first excited state.

1.Determine the probability of finding the particle in the right half of the hole at $t=0$.

My solution:

1) I first project the state into the coordinate representation:

$\psi(x) = \langle x|\psi \rangle = \frac{1}{\sqrt{2}}[\langle x|1\rangle + \langle x|2\rangle]$

I replace $\psi_1(x)$ and $\psi_2(x)$ in the formula, knowing that:

$\psi_1(x) = \sqrt{\frac {2}{a}} cos(\frac{\pi x}{a})$

$\psi_2(x) = \sqrt{\frac {2}{a}} sin(\frac{2\pi x}{a})$

I obtain:

$\psi(x) = \frac{1}{\sqrt{2}}[\sqrt{\frac {2}{a}} cos(\frac{\pi x}{a}) + \sqrt{\frac {2}{a}} sin(\frac{2\pi x}{a})]$

2) At this point to obtain the required probability I must integrate the framework module of $\psi(x)$. But $\psi(x)$ it's real. As a result I have:

$P_r = \int_0^{\frac{a}{2}} (\frac{1}{\sqrt{2}}[\sqrt{\frac {2}{a}} cos(\frac{\pi x}{a}) + \sqrt{\frac {2}{a}} sin(\frac{2\pi x}{a})])^2 dx$

I think the calculations I made are right. I get as a result: $P_r = \frac{3\pi a^2 + 4a^2}{24\pi}$

3) Now if i calculate the Prob for the left side of the potential hole i obtain: $P_l = \frac{3\pi a^2 - 4a^2}{24\pi}$

The sum has to be equal to 1, but: $P_{tot} = P_r + P_l= \frac{3\pi a^2 + 4a^2}{24\pi} + \frac{3\pi a^2 - 4a^2}{24\pi} = \frac {3a^2}{24}$

Either I did something wrong or my reasoning is wrong. Or I need to enter a normalization constant in $\psi(x)$ equal to the inverse root of this constant?

 

[hutchphd]

In step 2 your probability suddenly develops units of $a^2$. Ya might want to check that.
NB. the Hamiltonian is symmetric in x and so are all of the eigenstate probabilities. You needn't integrate anything explicitly, but it still will produce the correct answer when correctly done.
Good that you see there is an error! How did you get the minus sign in (3)???

Erratum: With thanks to @PeroK in #3 below. I was doing 65 in a 35 zone....apologies for error.

 

[PeroK]

I don't see how to do the problem without integration. Note that:
$$|\psi(x)|^2 = \frac 1 2 [|\psi_1(x)|^2 + |\psi_2(x)|^2] + \psi_1(x)\psi_2(x)$$You can use the symmetry of the first term to calculate that without integration. That leaves only the cross term (interference!).

PS same tip as yesterday. Stick with the general eigenfunctions as long as possible. At this level mathematics is much about the properties of things, rather than the specific things themselves.

 

[keyzan]

Ok so I can say that the probabilities of finding the particle in the entire hole in the states:
$P(1) = \frac{1}{2}$
$P(2) = \frac{1}{2}$

So the probability of finding it in the left or right half of the hole is:

$P(1) = \frac{1}{4}$
$P(2) = \frac{1}{4}$

(I find the same result by integrating)

We can finally find it in a combination of the two stationary states:
$P_{int} = \frac{2}{a}\int_0^\frac{a}{2}cos(\frac{\pi x}{a}) sin(\frac{2\pi x}{a})dx = \frac{8}{3\pi}$

I have many doubts:

1) Obviously the interference term is not a state (because a generic state is a linear combination of eigenstates). So how do I interpret that probability? The probability of finding the particle in an interference state?

2) I have reviewed the calculations several times. Why do I get that to the left or right of the hole the sum of each probability gives me almost 1? So the sum of the probabilities of finding the particle on the right and left is 2? I'm missing something. Maybe something I don't remember from the theory.

 

[PeroK]

You need to start from here:

$$|\psi(x)|^2 = \frac 1 2 [|\psi_1(x)|^2 + |\psi_2(x)|^2] + \psi_1(x)\psi_2(x)$$

Note that $\psi_1^2$ and $\psi_2^2$ are even functions. So, the integral of each from 0 to $+a/2$ must be $1/2$. That gives us:
$$P(>0) = \frac 1 2 + \frac 1 2\int_0^{a/2}\psi_1(x)\psi_2(x) \ dx$$Note also that the integrand in that interference term is an odd function. If we denote that whole term as $I$, then:
$$P(>0) = \frac 1 2 +I$$$$P(<0) = \frac 1 2 - I$$This not only ensures that the total probability is $1$, but should make sense from a physical perspective. Without the cross term, we would see a classical mixing of two symmetric states. Instead, we see a nonclassical interference term and an insight into the nature of QM.

Note finally that you may have missed the factor of $1/2$ in calculating $I$.

 

[PeroK]

Ok so I can say that the probabilities of finding the particle in the entire hole in the states:
$P(1) = \frac{1}{2}$
$P(2) = \frac{1}{2}$

So the probability of finding it in the left or right half of the hole is:

$P(1) = \frac{1}{4}$
$P(2) = \frac{1}{4}$

(I find the same result by integrating)

We can finally find it in a combination of the two stationary states:
$P_{int} = \frac{2}{a}\int_0^\frac{a}{2}cos(\frac{\pi x}{a}) sin(\frac{2\pi x}{a})dx = \frac{8}{3\pi}$

That's not a probability. That a part of the calculation of a probability. Quantum interference affects the probability, but there isn't a "probability of interference".

1) Obviously the interference term is not a state (because a generic state is a linear combination of eigenstates). So how do I interpret that probability? The probability of finding the particle in an interference state?

Definitely not. It's a calculation of the effect of interference in this case.

2)I have reviewed the calculations several times. Why do I get that to the left or right of the hole the sum of each probability gives me almost 1? So the sum of the probabilities of finding the particle on the right and left is 2? I'm missing something. Maybe something I don't remember from the theory.

Just a simple case of calculation error.

 

[hutchphd]

It might help to look at tthe time development of the mixed state
$|\psi \rangle = \frac{1}{\sqrt{2}}[|1\rangle + |2\rangle]$

In fact this is surely done on line. Found it: put the mixed state above into

https://www.st-andrews.ac.uk/physic.../SuperpositionStates/SuperpositionStates.html

So the "interference" terms provide the time dependence of the particle "sloshing" back and forth in the mixed state at the difference frequency. It is important to define exactly what the question is here. Makes you think about temporal coherence I guess. I like it.

 

[keyzan]

You're absolutely right, I forgot a 2. Everything is much clearer now.

Note also that the integrand in that interference term is an odd function. If we denote that whole term as $I$, then:
$$P(>0) = \frac 1 2 +I$$$$P(<0) = \frac 1 2 - I$$This not only ensures that the total probability is $1$, but should make sense from a physical perspective. Without the cross term, we would see a classical mixing of two symmetric states. Instead, we see a nonclassical interference term and an insight into the nature of QM.

So the probability of finding the particle to the left of the hole is very small compared to that of finding it to the right. I'll try to guess: this term, however, should oscillate when we insert the times, because, while in the square module the exponential does not bring any contribution (being a phase) in the interference term we will have an omega pulsation. Therefore the term I varies over time, the one just discussed being only a snapshot of the system at time t = 0.

 

[keyzan]

https://www.st-andrews.ac.uk/physic.../SuperpositionStates/SuperpositionStates.html

This is just spectacular

 

[PeroK]

You're absolutely right, I forgot a 2. Everything is much clearer now.



So the probability of finding the particle to the left of the hole is very small compared to that of finding it to the right. I'll try to guess: this term, however, should oscillate when we insert the times, because, while in the square module the exponential does not bring any contribution (being a phase) in the interference term we will have an omega pulsation. Therefore the term I varies over time, the one just discussed being only a snapshot of the system at time t = 0.

Yes, this a time-independent solution. As mentioned above, when you add the time-dependent factor, then the interference term, hence the probability oscillates with a frequency that depends on the difference in energy levels.

This is another important concept in QM. A superposition of two stationary states is not stationary.

 

[hutchphd]

Play with it!!

 

[keyzan]

Hi guys, today I would like to continue solving the exercise. The second point reads:

2. Determine the average value of the position at generic time t.
My solution:
I expand the following formula:

$\langle\psi\rangle = \langle\psi|\hat{X}|\psi\rangle = \frac{1}{2} \text{ }[ \langle 1| + \langle 2| ]\text{ } \hat{X} \text{ }[|1\rangle + |2\rangle] = \frac{1}{2} [\langle 1|\hat{X}| 1\rangle + \langle 2|\hat{X}| 2\rangle + \langle 1|\hat{X}| 2\rangle + \langle 2|\hat{X}| 1\rangle]$

Now I can replace the integrals:

$\langle\psi\rangle = \frac{1}{2} \text{ }[ \int_{\frac{a}{2}}^{- \frac{a}{2}} \psi_1(x) x \psi_1^*(x)dx + \int_{\frac{a}{2}}^{- \frac{a}{2}} \psi_2(x) x \psi_2^*(x)dx + \int_{\frac{a}{2}}^{- \frac{a}{2}} \psi_1(x) x \psi_2^*(x)dx + \int_{\frac{a}{2}}^{- \frac{a}{2}} \psi_2(x) x \psi_1^*(x)dx]$

Now I just have to solve the integrals.
Is the reasoning correct? Or are there also simpler and quicker methods without using integrals?

 

[PeroK]

You need the time dependent wavefunctions.

Note that the remaining inner product or integrals are complex conjugates of each other.

 

[hutchphd]

Note tthat the St Andrews site simulation will display this also on the graph. But work it out. You could also do this in the "Heisenberg Picture" and reproduce the Erhenfest Theorem result directly, but that is a little bit fraught. Your method is good (with explicit time dependence included)

 

[keyzan]

Ye I forgot to insert the variable t. So is $\psi(x,t)$. I studied Ehrenfest's theorem. I know that to go from the Shroedinger representation to the Heisenberg one (where the states do not evolve but the operators), you need to use the time evolution operator:

$\hat{A}_H(t) = \hat{U}^+(t, t_0) \hat{A}_S(t) \hat{U}(t, t_0)$

If I'm not mistaken then the time evolution operator was broadly defined as the exponential of the Hamiltonian.


So at this point I would have the average value of a time-dependent operator over states that are not time-dependent.

$\langle \hat{X} \rangle = \langle\psi_S(t_0)| \hat{X}_H(t) | \psi_S(t_0) \rangle$

Theoretically ok, but on a practical level how do you calculate something like this with Ehrenfest theorem?

 

[PeroK]

I'm on my phone, so it's difficult to type equation. You need the solution to full SDE:
$$\Psi(x,t) = \frac 1 {\sqrt 2}\bigg (\psi_1(x)\exp(-\frac{iE_1t}{\hbar}) + \dots \bigg)$$Then you calculate time dependent expected values.

 

[hutchphd]

In the end the result (of Heisenberg vs Schrodinger picture) will be the same (as it must). Even the algebra rapidly becomes the same (because the initial (t=0) condition is expanded in energy eigenstates from the beginning). It is probably a useful learning exercise to work it through both ways. Couldn't hurt.